EEVblog Electronics Community Forum
Electronics => Beginners => Topic started by: Legion on February 13, 2014, 07:06:08 pm
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I answered question 2 here: http://www.allaboutcircuits.com/worksheets/dc_p.html (http://www.allaboutcircuits.com/worksheets/dc_p.html)
The site's answer says the bulb shouldn't dim when the switch is closed since it's a parallel circuit and the voltage will remain the same (ignoring voltage sag).
However, wouldn't the current division caused by the parallel circuit result in a significant change in bulb brightness (assuming the resistance in series with the switch is not significantly higher than the bulb's resistance)? Wouldn't the drop in current through the filament have a much more significant effect than voltage sag?
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Nothing will happen to change the brightness because the resistor is in parallel so both get the same voltage and as much current as required. Now if for example a tiny wire was being used then it would cause a resistance and the more current drawn the mre voltage drop on the wire so yes the bulb would dim.
When i living in ital in a house with crappy wiring a plumber wanted to see if the electric boiler in the kitchen was working, so he turned on the kitchen (incandescent) light and then turned on and off the heater checking for a slight dimming of the light, this was because the wiring was so poor that a 10A load next to a 0.05A light bulb did make a difference but then i think question 2 is theoretical and not dealing with the practicalities of wire supplying the power
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Does this mean then that total circuit current will increase, but the current through the light bulb will remain constant?
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When i living in ital in a house with crappy wiring a plumber wanted to see if the electric boiler in the kitchen was working, so he turned on the kitchen (incandescent) light and then turned on and off the heater checking for a slight dimming of the light, this was because the wiring was so poor that a 10A load next to a 0.05A light bulb did make a difference but then i think question 2 is theoretical and not dealing with the practicalities of wire supplying the power
Does British wiring not do this? I would always expect flicking a 10A load on and off to visibly dim nearby lamps.
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I don't know, I've not used an electric heater near an incandescant bulb since i was back in the UK, the wiring in that particular house was really crap, it was like old 2 core speaker wire.
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Since nothing else is specified... with perfect wire and voltage source you could in theory run additional giga-tera-peta-amperes through the switch and the bulb would still be as bright as before.
As long as the voltage across the bulb is constant, the current will also remain constant.
If you consider that real wires and batteries have a resistance greater than zero, then switching in the resistance will have some effect. It's still plain old Ohms law, except there now some additional resistance that previously wasn't accounted for.